Which one of the following functions is one-one?

Options

A. `f:R to R" given by "f(x)|x-1|"for all "x in R`
B. `g:[-pi//2,pi//2] in R` is given by:
`g(x)=|sin x|"for all "x in[-pi//2,pi//2]`
C. `h:[-pi//2,pi//2] in R` is given by
`h=(x)=sin x" for all "x in [-pi//2, pi//2]`
D. `phi: R to R "given by" f(x)=x^(2)-4"for all x " inR`

Explanation

To determine which of the given functions is one-one, we will analyze each option step by step.

### Step 1: Understand the Definition of One-One Function
A function \( f \) is called one-one (or injective) if for every pair of distinct inputs \( a \) and \( b \) in the domain, the outputs are also distinct. In mathematical terms, if \( f(a) = f(b) \), then it must follow that \( a = b \).

### Step 2: Analyze Each Option

#### Option 1: \( f(x) = |x| - 1 \)
- **Check for one-one property**:
- Let's evaluate \( f(-1) \) and \( f(3) \):
- \( f(-1) = |-1| - 1 = 1 - 1 = 0 \)
- \( f(3) = |3| - 1 = 3 - 1 = 2 \)
- Since \( f(-1) \neq f(3) \), we need to check more values:
- \( f(1) = |1| - 1 = 0 \)
- \( f(-1) = 0 \)
- Here, \( f(-1) = f(1) \) but \(-1 \neq 1\), so this function is not one-one.

#### Option 2: \( g(x) = |\sin x| \) for \( x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \)
- **Check for one-one property**:
- Evaluate \( g(-\frac{\pi}{2}) \) and \( g(\frac{\pi}{2}) \):
- \( g(-\frac{\pi}{2}) = |\sin(-\frac{\pi}{2})| = | -1 | = 1 \)
- \( g(\frac{\pi}{2}) = |\sin(\frac{\pi}{2})| = | 1 | = 1 \)
- Since \( g(-\frac{\pi}{2}) = g(\frac{\pi}{2}) \) but \(-\frac{\pi}{2} \neq \frac{\pi}{2}\), this function is also not one-one.

#### Option 3: \( h(x) = \sin x \) for \( x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \)
- **Check for one-one property**:
- The sine function is strictly increasing in the interval \([- \frac{\pi}{2}, \frac{\pi}{2}]\).
- If \( h(a) = h(b) \) implies \( \sin a = \sin b \), then \( a = b \) in this interval.
- Therefore, this function is one-one.

#### Option 4: \( k(x) = x^2 - 4 \)
- **Check for one-one property**:
- Evaluate \( k(-1) \) and \( k(1) \):
- \( k(-1) = (-1)^2 - 4 = 1 - 4 = -3 \)
- \( k(1) = (1)^2 - 4 = 1 - 4 = -3 \)
- Since \( k(-1) = k(1) \) but \(-1 \neq 1\), this function is not one-one.

### Conclusion
After analyzing all options, we find that only **Option 3** \( h(x) = \sin x \) for \( x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \) is a one-one function.